Metallic Ball

Algebra Level 2

When a metallic ball bearing is placed inside a cylindrical container, of radius 2 cm, the height of the water, inside the container, increases by 0.6 cm. The radius, to the nearest tenth of a centimeter, of the ball bearing is -


The answer is 1.2.

Karan Ahire by Karan Ahire , 21, India

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1 solution

The volume of the bearing is equal to the volume of the rise of water. Let r r be the radius of the bearing. The volume of the bearing is 4 3 π r 3 \dfrac{4}{3}\pi r^3 . Let R R be the radius of the container. The volume of the rise of water is π R 2 × h = π ( 2 2 ) ( 0.6 ) = 2.4 π \pi R^2 \times h=\pi (2^2)(0.6)=2.4 \pi . Since the two volumes are equal, we equate them

4 3 π r 3 = 2.4 π \dfrac{4}{3} \pi r^3 = 2.4 \pi

r ≈ 1.2 r \approx \boxed{1.2}

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